# Group matrix values by column [closed]

Original Question:
I have a list {{a, 1}, {b, 3}, {c, 1}, {d, 2}, {e, 3}}. I want to be able to group the the first number in each pair that has the same second number in the pair. So i would get a list like

{{a,c},d,{b,e}} so

{a,c} is from {a,1} and {c,1}

d is from {d,2}

{b,e} is from {b,3} and {e,3}

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I think I know what you need:

SortBy[GatherBy[list,Last],Last@*Last][[;;,;;,1]]/.{x_?AtomQ}->x

The result is:

{{a,c},d,{b,e}}

This will gather the elements by the last member of each small list, sort it by the index number and then throw away all the index numbers. Finally, it can change all one element list into the number itself.

Will this help directly?

@HighPerformanceMark oh yes! I mistype it cause I’m currently typing with my tiny phone on my hand. Thanks for informing!
– Wjx
Jun 17 at 7:32

I am not sure I understand. Here are some ways to group by last element:

list={{1, 1}, {1, 3}, {2, 1}, {2, 2}, {2, 3}}
GatherBy[list, Last]
GroupBy[list, Last]
Last@Reap[Sow[{##}, #2] & @@@ list, _, Rule]

yielding respectively:

{{{1, 1}, {2, 1}}, {{1, 3}, {2, 3}}, {{2, 2}}}

<|1 -> {{1, 1}, {2, 1}}, 3 -> {{1, 3}, {2, 3}}, 2 -> {{2, 2}}|>

{1 -> {{1, 1}, {2, 1}}, 3 -> {{1, 3}, {2, 3}}, 2 -> {{2, 2}}}

GroupBy has other useful features.

1

Probably GroupBy[list, Last -> First] with skiping {} from one element lists. Not so sure about an order and its meaning.
– Kuba
Jun 17 at 7:21

I don’t mind the extra {}. But how do i use this as a list it has the weird form <|1 -> {1, 2}, 3 -> {1, 2}, 2 -> {2}|>
– MTR
Jun 17 at 7:28

@MTR that is an Association (see documentation), You can “normalize” it with Normal@GroupBy[list,Last].
– ubpdqn
Jun 17 at 7:31

@MTR or Values @ GroupBy[list, Last -> First]
– Kuba
Jun 17 at 7:34